Hello there,

I thought i would share this square roots trick. It allows calculating the square root of somewhat large numbers like 9801.

In fact, i'll show the solution using that number to start out, 9801.

First we make a two column table of squares from 1 to 9:

1, 1

2, 4

3, 9

4, 16

5, 25

6, 36

7, 49

8, 64

9, 81

Next we add a third column which is simply the last digit of all the squares (so 16 becomes 6):

1, 1, 1

2, 4, 4

3, 9, 9

4, 16, 6

5, 25, 5

6, 36, 6

7, 49, 9

8, 64, 4

9, 81, 1

Ok now we are ready to begin finding the square root of 9801 but we use the table for any number.

First we look at the last digit, which is a 1 in this case.

Now we look at the third column above and find that a '1' appears as the squares of both 1 and 9. Remember that 1 and 9.

Next, we look at the first two digits which are 98. We find in the table a square that is less than 98. This has to be 81 which is the square of 9. This means the first number of the square root is 9.

Next, we multiply that 9 times the next integer which is 10, and get 90. Now 98 (first two digits again) is greater than 90, so then looking at that 1 and 9 from above again we choose the 9 because it is greater than 1, and thus the second number is 9.

The total solution then is 99, and that is the square root of 9801.

Now we will find the square root of 1369.

First we find the last digit the '9' appears in third column of the table for squares of both 3 and 7 so keep those two in mind.

Next looking at the first two digits, the square that is just below 13 is 9, which is the square of 3, so 3 is the first digit of the solution.

Next, 3*4=12, and 13 is greater than 12, so out of the two we got above (the 3 and the 7). we choose the 7 because it is greater than 3.

So the final result is 37 and that's the square root of 1369.

This gets a bit more tricky with squares that are not perfect squares. For say 1370 we'd have to make a choice as to what number we want to start with for the last two digits so this is mostly for perfect squares.

I believe this works because square roots are related to the grouping of every 2 digits of a number, whole or fractional, and the way integers only combine into a limited number of solutions.

I thought i would share this square roots trick. It allows calculating the square root of somewhat large numbers like 9801.

In fact, i'll show the solution using that number to start out, 9801.

First we make a two column table of squares from 1 to 9:

1, 1

2, 4

3, 9

4, 16

5, 25

6, 36

7, 49

8, 64

9, 81

Next we add a third column which is simply the last digit of all the squares (so 16 becomes 6):

1, 1, 1

2, 4, 4

3, 9, 9

4, 16, 6

5, 25, 5

6, 36, 6

7, 49, 9

8, 64, 4

9, 81, 1

Ok now we are ready to begin finding the square root of 9801 but we use the table for any number.

First we look at the last digit, which is a 1 in this case.

Now we look at the third column above and find that a '1' appears as the squares of both 1 and 9. Remember that 1 and 9.

Next, we look at the first two digits which are 98. We find in the table a square that is less than 98. This has to be 81 which is the square of 9. This means the first number of the square root is 9.

Next, we multiply that 9 times the next integer which is 10, and get 90. Now 98 (first two digits again) is greater than 90, so then looking at that 1 and 9 from above again we choose the 9 because it is greater than 1, and thus the second number is 9.

The total solution then is 99, and that is the square root of 9801.

Now we will find the square root of 1369.

First we find the last digit the '9' appears in third column of the table for squares of both 3 and 7 so keep those two in mind.

Next looking at the first two digits, the square that is just below 13 is 9, which is the square of 3, so 3 is the first digit of the solution.

Next, 3*4=12, and 13 is greater than 12, so out of the two we got above (the 3 and the 7). we choose the 7 because it is greater than 3.

So the final result is 37 and that's the square root of 1369.

This gets a bit more tricky with squares that are not perfect squares. For say 1370 we'd have to make a choice as to what number we want to start with for the last two digits so this is mostly for perfect squares.

I believe this works because square roots are related to the grouping of every 2 digits of a number, whole or fractional, and the way integers only combine into a limited number of solutions.

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